Chapter 10, Problem 10.121QA

Interpretation Introduction

To identify:

The element $X$ from the reaction of ammonia and unknown acid $HX$ by using $1.0 m$ glass tube.

Answer to Problem 10.121QA

Solution:

The Element $X$ is Bromine ($Br$)$.$

Explanation of Solution

1) Concept:

Graham’s law of diffusion states that the diffusion rate of a gas is inversely proportional to the square root of its molar mass.We can use the equation of Graham’s law of diffusion to find the atomic mass of element $X$ and from its molar mass we can identify it the element $X$.

In this reaction, ammonia and $HX$ are $1.00 m$ away from each other in a glass tube. When ammonia reacts with $HX$ acid it forms a product at $68.5 m$.

2) Formula:

$\frac{{effusion rate}_{{NH}_3}}{{effusion rate}_{HX}}= \sqrt{\frac{{molar mass}_{HX}}{{molar mass}_{{NH}_3}}}$

3) Given:

i) $lenght of glass tube=1.00 m=100 cm$.

ii) $Diffusion of N{H}_3=68.5cm$.

iii) $\mathcal{M}\mathcal{ }of N{H}_3=17.0\frac{g}{mol}$

4) Calculations:

The molar mass of the ammonia is $17.0 g/mol$. So, the diffusion rate of ammonia is greater than HX acid. So, the distance travelled by the ammonia gas is $68.5 cm$.

Calculate the distance travelled of the $HX$ acid in the tube.

Diffusion of $HX=100 cm-68.5 cm=31.5 cm$

$\frac{{effusion rate}_{{NH}_3}}{{effusion rate}_{HX}}= \sqrt{\frac{{molar mass}_{HX}}{{molar mass}_{{NH}_3}}}$

$\frac{68.5 cm}{31.5 cm}=\sqrt{\frac{{molar mass}_{HX}}{17.03 \frac{g}{mol}}}$

$2.1746=\sqrt{\frac{{molar mass}_{HX}}{17.03 \frac{g}{mol}}}$

Now take the square of both sides.

$4.7289=\frac{{molar mass}_{HX}}{17.03 \frac{g}{mol}}$

${molar mass}_{HX}=4.7289\times 17.03 \frac{g}{mol}=80.5329\frac{g}{mol}$

${molar mass}_{HX}=atomic mass of H+atomic mass of X$

$80.5329\frac{g}{mol}=1.0 \frac{g}{mol}+atomic mass of X$

$atomic mass of X=80.5329\frac{g}{mol}-1.0\frac{g}{mol}=79.53\frac{g}{mol}$

$79.53\frac{g}{mol}$ is the atomic mass of Bromine.

Therefore, the unknown element $X$ in this reaction is Bromine $\left(Br\right)$.

Conclusion:

Using diffusions of ammonia and $HX$ acid, we got the atomic mass of $X$. From the atomic mass, we identified the element is Bromine.